Following on from the under-whelming success of my recent SAND+SUN+SEX+SEA=IBIZA alphametic puzzle and its solution, I recalled another exercise in mathematics from earlier this year. One of my granddaughters sent me a text message asking if I could solve the following 3-variable set of linear simultaneous algebraic equations.
x+2y+3z = 14 (Eqn. 1)
2x+3y+4z = 20 (Eqn. 2)
5x+5y+5z = 30 (Eqn. 3)
Easy-peasy I thought. I enjoy algebra and this would be a doddle—the sort of thing you would have done in year 8 or 9. A couple of substitutions and out would pop the answer. So, I set about substituting, expanding and tidying up and singularly failed to solve the equations. Watch…
From Eqn. 1, x = 14 – 2y – 3z.
Substituting for x in Eqn. 2, 2(14 – 2y – 3z) +3y + 4z = 20 which tidies to y + 2z = 8 (Eqn. 4)
Substituting for x in Eqn. 3, 5(14 – 2y – 3z) +5y + 5z = 30 which tidies to, guess what, y + 2z = 8 (Eqn. 4 again)
Now what? I scratched my head, chewed the end of my pen, stared out of the window and generally waited for inspiration. I was particularly intrigued by the fact that there was a solution. Apparently the mathematics’ teacher at my granddaughter’s school had asked the pupils to derive the solution x = 1, y = 2, z = 3. This solution works. Try it.
Gabriel Cramer, Swiss mathematician, 1704 – 1752
Finally, I remembered from the dim distant past (the mid-60s in my case) that there was a plug-in-the-numbers-and-turn-the-handle method based on determinants. Jeez, I could barely spell determinants let alone recall what they were. I retrieved one of my degree-course books on mathematics and there it was—Cramer’s Rule. Yes! All I had to do was to remember how to apply it. That took a bit of deep memory retrieval, aided by my maths book and a very useful website, plus a few cups of coffee but I did it. Again, if you are interested to see the workings, send me an e-mail and I’ll send you my solution, no strings attached.
The bottom line is that applying Cramer’s Rule did not come up with a solution. For those who know about this rule, the denominator determinant, D, given by the coefficients of x, y and z computes to 0. This is bad news as D is a denominator in the final calculations—that is:
x = DX/D
y = DY/D
z = DZ/D
where DX = the x-numerator determinant, DY = the y-numerator determinant, and DZ = the z-numerator determinant.
In all cases, dividing by zero produces an undefined result. When this happens using Cramer’s Rule, the set of equations is termed inconsistent or indeterminate meaning the answer cannot be determined.
So, how did the maths teacher produce the x = 1, y = 2, z = 3 result? I asked my granddaughter to bring me home a copy of the teacher’s workings but, apparently, all he did was state the result without proof and then move on to another problem. Hmm. How did he do it? By trial and error? And why did Cramer’s Rule fail? To this day, I don’t know the answers to these questions.
By the way, a couple of comments on my alphametic posting suggested I am bored and have nothing better to do than work on mathematical problems which, by implication, is also boring. Nothing could be further from the truth. First, I am not bored. I have plenty of things to keep me occupied in my retirement. Second, I have never found mathematics to be boring—tough sometimes, perplexing other times but never boring. Third, the ex-engineer in me rises to the challenge of solving the problem. And fourth, the ex-teacher in me rises to the challenge of presenting a solution that is understandable even if mathematics is not your thing. It’s interesting that those who claim I am bored never asked for my solution. What can you infer from that?